Communication Systems (ECE 301): Analysis of Sample Exam Questions

This document presents a detailed analysis of sample exam questions covering various aspects of communication systems. The questions address topics such as the advantages and disadvantages of fiber optics and copper wires, the characteristics of digital signals, and the challenges of mobile network connectivity. Further, the document explores the average quantization error in A-D converters, minimum sample rates for audio codecs, and frequency-shift keying. It delves into the workings of synchronous TDM and asynchronous transmission, along with data compression techniques, including run length encoding and index compression. The document also touches on asymmetric cryptography, CSMA/CA, and the concept of ISMA. Finally, it includes calculations for maximum data carrying capacity and discusses the application of these concepts in real-world scenarios.

Sample exams questions 1
SAMPLE EXAMS QUESTIONS
By Name
Course
Instructor
Institution
Location
Date

Sample exams questions 2
Question 1 a.
a).
Advantages of fiber-optics over copper in communication media.
1. Greater bandwidth
Fiber –optics have relatively more bandwidths as compared to copper. This is because copper
wires were originally designed for a voice transmission hence lower bandwidth. Fiber optic
wires offer higher bandwidth for conveying higher data than copper wires of the equal sizes.
2. Reduced Security Risk:
The progression of the fiber optic communiqué sale is essentially compelled by aggregate
alertness in the data security anxieties and employment of the substitute raw material.
3. Longer Distance:
For the fiber optic broadcast, optical wires are able of offering less power loss that permits
signals to be conveyed to a more distance than for the copper wires.
4. Noise:
Practically transmission wires utilization, it’s unavoidable to achieve environments parameters
such as heating, ventilating, power substations among other industrialized sources of noise.
Nevertheless, fiber has a smaller rate of bit error of 10×10-13, as an effect of fiber of very
resilient to electromagnetic disturbances of transmission of Fiber optic is hence essentially
noise free as compared to copper.
Advantages of copper over fiber-optics.

Sample exams questions 3
5. The Cost is higher than in Copper wires:
Notwithstanding the truth is that optic fiber connection costs are falling by around 60% yearly,
connecting optic fiber wiring is comparatively higher than the cables for copper. This is true
since copper cable connection doesn’t require additional care like fiber cables.
b)
Digital signals are a sequences of pulsates with two states - on (illustrated by the logic ‘1’) or off
(illustrated by the logic ‘0’). Digital signals convey more data per second as compared to
analogue signals and hence they keep their quality of signal better over long distances of
transmission of signal. Therefore, digital signals are better in transmission of signal as the noise
is relatively lower as compared to the analogue signal as seen in the diagram below.
Fig 1: Showing the noise signal in the digital signal.
c)

Sample exams questions 4
1. It is really annoying as the mobile may indicate "bars" of network and indicate a good
connection of network, yet it is still incapable to do some very significant tasks.
2. Disappointingly, the Droid X hotspot and, to a slighter level, the 4G hotspot Epic were
not that easy to use.
d)
The channel connection in GSM between two mobile subscribers simplex is a full duplex since
for this system, these parties are capable to connect with each other concurrently. A prototype of
a full-duplex device is a cell phone; the individual at the two ends of a call are able to speak and
can be heard by the other individual concurrently (Max, 2014).
Question 2a
A)
The average quantization error in an A-D converter is ∆V/4 since during the sampling stage, a
full wave which has 2 cycles is used in the conversion. These two cycles will have 4 voltage
point hence the error can easily be obtained through ∆V/4.
b)

Sample exams questions 5
Minimum sample rates is 480 Hz for the A Codec for a mobile phone is required to convert
speech in the range 400Hz- 2kHz into a digital data stream.
C)
Frequency-shift keying permits digital data to be conveyed by variations or swings in the
frequency of a carrier signal, most normally it is an analog sine wave which is also a carrier
wave. There are always 2 binary states in a signal, logic one (1) and logic zero (0), each of these
are represented by an analog wave form (Johns, 2013).
d)
Synchronous TDM operates through the muliplexor hence resulting to accurately the equal
amount of time to every device connected to it. This duration of time portion is allotted even if a
device doesn’t have something to transmit. This is uneconomical since there will be several time
durations if the allocated time apertures are not being employed. While Time-division

Sample exams questions 6
multiplexing is a technique of placing several data streams in a sole signal through unraveling
the signal into several segments, every one has a very short duration of time.
Advantages of Synchronous
This reduces overhead bits
Advantages of adaptive time division multiplexing
TDM circuitry is not very complex
Question 3 a
a)
It helps in solving the problem of little frequency range (Bandwidth). This is the punishment for
acquaint with transitions of frequency. For a LAN of 10 Mbps, the spectrum signal which is
between the 5 and 20 MHz. The data word 111001 would look as the below;
b)
g/Symbol Frequency of Occurrence

Sample exams questions 7
[per 120 symbols]
# 34
@ 28
& 22
% 16
£ 12
$ 8
34 28 '@' 22 '&' 16 '%' 12 '£' 8 '$'
/ \
2 'b' 7 'c'

Sample exams questions 8
Now, repeat this process until only one tree is left:
34
/ \
9 12 'a' 13 'd' 14 'e' 8 'f'
/ \
2 'b' 7 'c'
21 27
/ \ / \
9 12 'a' 13 'd' 14 'e' 85 'f'
/ \
2 'b' 7 'c'
48
/ \

Sample exams questions 9
21 27
/ \ / \
9 12 'a' 13 'd' 14 'e' 85 'f'
/ \
2 'b' 7 'c'
133
/ \
48 85 'f'
/ \
21 27
/ \ / \
9 12 'a' 13 'd' 14 'e'
/ \
2 'b' 7 'c'
C)
c) Actions performed by Sender and Receiver
In cryptography of asymmetric, there are receiver and sender and each of them contain a key set
which is remote and the other one is public. The actions performed by both of them include;

Sample exams questions 10
Private-public keys
The sender will encrypt the message using its remote key when he wants to send a message then
he uses the communal key from the encrypts and receiver o'er.
Message Encrypted
The message which was encrypted is then conveyed to the receiver. The receiver uses its private
key to decrypt the encryption of outside on the message and then senders public key to decrypt
the message which was encrypted with the sender’s private key.
Message Decrypted
The original message is received by the receiver and this is how asymmetric cryptography in
different ways.
Digital Signature
The receiver gets the original document and is capable of proving that the sender since
unlocking of the message is done by public keys from the sender and this proves that the identity
of the person who sends the message.
d)
During the early 1970s there was key public progresses. The publication of the draft Data
Encryption Standard which was undertaken in the U.S. Federal Register on 17 March 1975. The

Sample exams questions 11
publication and the effort of coming up with a secure electronic communication demanded a lot
of finances which was not available at that time
Question 4a
a)
P-Persistent Carrier-sense multiple access is a media access control (MAC) protocol where a
node confirms the nonappearance of other traffic afore conveying on a shared transmission
medium like an electrical bus
b)
i. Gauge min and max values
Example
Taking the example of age as a variable, the aim should be to ensure that my minimum value is
sensible. Additionally, maxing should not be done somewhere in the region of 400’s.
ii. Look for missings.
Example
Columns are sorted in both ascending and descending order to identify the missing values in the
columns. Additionally, one can filter their dataset to only look at records that have a missing
value.
iii. Check the values of categorical variables
Example:

Sample exams questions 12
Correcting the categories of a spreadsheet from the upper part down to the lower part thus
making it easier to detect errors early enough
iv. Look at the ‘incidence rate’ of binary variables
Example,
Having 1’s and nulls instead of 1’s and 0’s which is easier to identify since the rate of the binary
variable will equate to 1.
c)
Question 5a
a) Calculating the maximum data carrying capacity of the cable
C= B log (1 + SNR)
C=?
B= 150 MHz
SNR= 1000
C= B log (1 + SNR)